evaluate exponential using Euler identity

Question

let us consider following exponential

$e^{-j*\pi*k/2}$ and $e^{j*\pi*k/2}$

we can decompose it as $cos(\pi*k/2)-j*sin(\pi*k/2)$ and second one same with plus sign

$cos(\pi*k/2)+j*sin(\pi*k/2)$

now for $k$ integer,first one is equal to $-j$ and second $j$ right?thanks in advance

Answer 1

No, for first one the pattern is $-i,-1,i,1,...$ and for second one the pattern is $i,-1,-i,1,..$

Answer 2

Not quite. The values of (using your "$j$") notation $e^{j\pi k/2}$ cycle among $1$, $j$, $-1$, and $-j$ in order as $k$ moves through the integers $0,1,2,\ldots$. They cycle in reverse order as you move backwards through the negative integers, which gives you the values for the other expression you wrote.

Said differently, you can write $$e^{\pm j\pi(4n)/2}=1$$ $$e^{\pm j\pi(4n+1)/2}=\pm j$$ $$e^{\pm j\pi(4n+2)/2}=-1$$ $$e^{\pm j\pi(4n+3)/2}=\mp j$$ for integral $n$.